西要毛子科技大学函数的求导法则XIDIANUNIVERSITYdy例5设=Incos(e'),求dxdy = [ln cos(e')]解dx1[cos(e'")]cos(e*)1cos(er) [-sin(e') (e*)= -e* tan(e'").函数的求导法则 例5 设 lncos(e ) 求  x y = ， d d y x 解 1 cos(e )x =  1 cos(e )x =  e tan(e ). x x = − d d y x = [ln cos(e )] x  [cos(e )] x  [ sin(e )] x − (e )x  